Answer
The solution set is {$\frac{3-\sqrt {5}}{4}, \frac{3+\sqrt {5}}{4}$}.
Work Step by Step
$x^{-2} - 6x^{-1} = -4$
Let $u=x^{−1}$
Rewrite the given equation as a quadratic equation in $u$ and solve for $u$.
$x^{-2} - 6x^{-1} = -4$: $u^2 - 6u = -4$
Solve for $u$ by completing the square.
Add $(\frac{-6}{2})^2 = 9$ to both sides.
$u^2 - 6u +9= -4+9$
$u^2 - 6u +9= 5$
$(u -3)^2= 5$
Apply the square root property.
$u -3= ±\sqrt {5}$
$u = 3±\sqrt {5}$
Solve for $u$.
$u=x^{−1}$
$3±\sqrt {5}=x^{−1}$
$3±\sqrt {5}=\frac{1}{x}$
$x= \frac{1}{3±\sqrt {5}}$
$x= \frac{1}{3+\sqrt {5}}$ or $x= \frac{1}{3-\sqrt {5}}$
$x= \frac{3-\sqrt {5}}{4}$ or $x= \frac{3+\sqrt {5}}{4}$
The solution set is {$\frac{3-\sqrt {5}}{4}, \frac{3+\sqrt {5}}{4}$}.
